of 1 44
The Mathematical Nature Of AI and Life
By
Ian Beardsley
Copyright © 2021 by Ian Beardsley
of 2 44
Contents
Abstract……………………………..3
Important……………….…………..4
The Computation……….…………..5
The Dynamic Function……………..5
Free Electron Model………………..13
Band Theory……………………….14
Silicon And Carbon……..…………15
Germanium and Carbon………….17
Fundamental AI Bio Eqns…………19
Using Fundamenta Eqns………….20
Conclusion…………………………22
Semiconductor Devices……………24
The Mathematical Connection……26
The Broad Picture………………….30
Divergence………………………….31
The Heart of the Matter…………..39
of 3 44
Abstract
I am finding mathematical connections between artificial intelligence and biological life and presenting
them as mathematical constructs. Of course if the structures of such elements are mathematical they can
be thought of as tensors making them invariant under transformation and thus Natural Law. However I
then proceed to to show what this mathematical structure is in terms of the modeling of such materials
with quantum chemistry to explain it. The QM result seems to make sense intuitively but much work
remains to be done. The proposition is speculative but compelling.
of 4 44
Important
Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the elements. As
you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and germanium are in
group 14 meaning they have 4 valence electrons and want 4 for more to attain noble gas electron
configuration. If we dope Si with B from group 13 it gets three of the four electrons and thus has a
deficiency becoming positive type silicon and thus conducts. If we dope the Si with P from group 15 it
has an extra electron and thus conducts as well. If we join the two types of silicon we have a
semiconductor for making diodes and transistors from which we can make logic circuits for AI.
As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the right of Si
but doping agent B is not directly to the left, aluminum Al is. This becomes important. I call (As-Ga) the
differential across Ge, and (P-Al) the differential across Si and call Al a dummy in the differential because
boron B is actually used to make positive type silicon.
That the AI elements make a three by three matrix they can be organized with the letter E with subscripts
that tell what element it is and it properties, I have done this:
Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the first row
and has 4 valence electrons making it carbon (C). I believe that the AI elements can be organized in a 3 by
3 matrix makes them pivotal to structure in the Universe because we live in three dimensional space so
the mechanics of the realm we experience are described by such a matrix, for example the cross product.
Hence this paper where I show AI and biological life are mathematical constructs and described in terms
of one another.
We see, if we include the two biological elements in the matrix (E14) and and (E15) which are carbon and
nitrogen respectively, there is every reason to proceed with this paper if the idea is to show not only are
the AI elements and biological elements mathematical constructs, they are described in terms of one
another. We see this because the first row is ( B, C, N) and these happen to be the only elements that are
not core AI elements in the matrix, except boron (B) which is out of place, and aluminum (Al) as we will
see if a dummy representative, makes for a mathematical construct, the harmonic mean. Which means we
have proved our case because the first row if we take the cross product between the second and third rows
are, its respective unit vectors for the components, meaning they describe them!
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
of 5 44
The Computation
And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological elements C
and N being the unit vectors are multiplied by the AI elements, meaning they describe them! But we have
to ask; Why does the first row have boron in it which is not a core biological element, but is a core AI
element? The answer is that boron is the one AI element that is out of place, that is, aluminum is in its
place. But we see this has a dynamic function.
The Dynamic Function
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon (Si) and
germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic (As) have an
asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C) and as such have 4
valence electrons. Thus to have positive type silicon and germanium, they need doping agents from group
13 (three valence electrons) like boron and gallium, and to have negative type silicon and germanium they
need doping agents from group 15 like phosphorus and arsenic. But where gallium and arsenic are in the
same period as germanium, boron is in a different period than silicon (period 2) while phosphorus is not
(period 3). Thus aluminum (Al) is in boron’s place. This results in an interesting equation.
A = (Al, Si, P )
B = (G a, G e, As)
A ×
B =
B
C
N
Al Si P
G a Ge As
= (Si ⋅ As − P ⋅ G e)
B + (P ⋅ G a − Al ⋅ As)
C + (Al ⋅ G e − Si ⋅ G a)
N
A = 26.98
2
+ 28.09
2
+ 30.97
2
= 50g /m ol
B = 69.72
2
+ 72.64
2
+ 74.92
2
= 126g /m ol
A ⋅
B = A Bcosθ
cosθ =
6241
6300
= 0.99
θ = 8
∘
A ×
B = A Bsi nθ = (50)(126)sin8
∘
= 877.79
877.79 = 29.6g /m ol ≈ Si = 28.09g /m ol
Si(A s − G a) + G e(P − Al )
SiG e
=
2B
Ge + Si
of 6 44
The differential across germanium crossed with silicon plus the differential across silicon crossed with
germanium normalized by the product between silicon and germanium is equal to the boron divided by
the average between the germanium and the silicon. The equation has nearly 100% accuracy (note: using
an older value for Ge here, is now 72.64 but that makes the equation have a higher accuracy):
Due to an asymmetry in the periodic table of the elements due to boron we have the harmonic
mean between the semiconductor elements (by molar mass):
This is Stokes Theorem if we approximate the harmonic mean with the arithmetic mean:
We can make this into two integrals:
If in the equation (The accurate harmonic mean form):
We make the approximation
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(As − G a) +
Ge
B
(P − Al ) =
2SiG e
Si + G e
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
∫
1
0
∫
1
0
[
Si
B
(As − G a) +
Ge
B
(P − Al )
]
d xd y ≈
1
Ge − Si
∫
Ge
Si
x d x
∫
1
0
∫
1
0
Si
B
(As − G a)d yd z ≈
1
3
1
(Ge − Si )
∫
Ge
Si
x d x
∫
1
0
∫
1
0
Ge
B
(P − Al ) d x dz ≈
2
3
1
(Ge − Si )
∫
Ge
Si
yd y
Si
B
(As − G a) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
2SiGe
Si + Ge
≈ Ge − Si
of 7 44
Then the Stokes form of the equation becomes
Thus we see for this approximation there are two integrals as well:
For which the respective paths are
One of the double integrals on the left is evaluated in moles per grams, the other grams per mole
(0 to 1 moles per gram and 0 to 1 grams per mole).
The Geometric Interpretation…
∫
1
0
∫
1
0
[
Si
B
(As − G a) +
Ge
B
(P − Al )
]
d yd z =
∫
Ge
Si
d x
∫
1
0
∫
1
0
Si
B
(As − G a)d yd z =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al ) d ydz =
2
3
∫
Ge
Si
dz
y
1
=
1
3
B
SiGa
ln(z)
y
2
=
2
3
B
Si Al
ln(z)
of 8 44
of 9 44
By making the approximation
In
We have
is the differential across Si, is the differential across
Ge is the vertical differential."
Which is Ampere’s Circuit Law
We see if written
Which is interesting because it is semiconductor elements by molar mass, which are used to
make circuits.
We say (Phi) is given by
and
And
=1.618
=0.618
(phi) the golden ratio conjugate. We also find
2SiGe
Si + Ge
≈ Ge − Si
Si(As − G a)
B
+
Ge(P − Al )
B
=
2SiGe
Si + Ge
Si
ΔGe
ΔS
+ G e
ΔSi
ΔS
= B
ΔSi = P − Al
ΔGe = As − Ga
ΔS = Ge − Si
∇ ×
B = μ
0
J + μ
0
ϵ
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
Φ
a = b + c
a
b
=
b
c
Φ = a /b
ϕ = b /a
ϕ
(ϕ)ΔGe + (Φ)ΔSi = B
of 10 44
Thus since
And we have
We see and are both and c is in the Si (silicon) field wave, but for E and B fields c is the speed of
light.
To find the Si wave our differentials are
∇ ×
B = μ
J + μϵ
0
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
ΔGe =
ΔS
Si
B −
Ge
Si
ΔSi
(
∇
2
−
1
c
2
∂
2
∂t
)
E = 0
(
∇
2
−
1
c
2
∂
2
∂t
)
B = 0
c =
1
ϵ
0
μ
≈ ϕ
μ
ϵ
0
Φ
ϕ
ϵ
0
= 8.854E − 12F ⋅ m
−1
μ = 1.256E − 6H /m
Ge
Si
= μϵ
0
ΔS
Si
= μ
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
Si = 0
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
Ge = 0
of 11 44
It is amazing how accurately we can fit these differentials with and exponential equation for the upward
increase. The equation is
This is the halfwave:
ΔC = N − B = 14.01 − 10.81 = 3.2
Δ Si = P − Al = 30.97 − 26.98 = 3.99
ΔG e = A s − G a = 74.92 − 69.72 = 5.2
Δ Sn = Bi − In = 121.75 − 114.82 = 6.93
ΔPb = Bi − T l = 208.98 − 204.38 = 4.6
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
of 12 44
Interestingly, the 0.4 is boron (B) over aluminum (Al) the very two elements that lead us to looking for a
wave equation because boron was the out of place element in the AI periodic table that lead to us using
aluminum as its dummy representative in the Si differential and that itself divided into the left hand terms
to give us the harmonic mean between the central AI elements semiconductor materials Si and Ge. The
Ag and Cu are the central malleable, ductile, and conductive metals used in making electrical wires to
carry a current in AI circuitry.
y(x) = e
B
Al
x
+
Ag
Cu
B
Al
=
10.81
26.98
= 0.400667
Ag
Cu
=
107.87
63.55
= 1.6974 ≈ 1.7
of 13 44
Free Electron Model
Metals are held together by free electrons that move throughout the solid. The free electron
model views these electrons as a gas. For the one-dimensional case a line through which the
electrons move and have no collisions with one another which is a one-dimensional square well
whose walls are the edge of a rod and respects the exclusion principle; no two electrons in an
orbit can have the same state, there can only be spin up and spin down. For the special case
where its temperature T is 0 degrees K the the energy levels are filled with electrons from the
from lowest to highest two at a time until the highest energy level is filled. This highest energy
level filled is the Fermi energy . Moving to the three dimensional case with electrons moving
freely in a metal block we consider it a three-dimensional infinite square well and solve the
Schrodinger equation:"
"
Where the wave functions of the electron states are:"
"
Three L’s with sides x,y,z, and three n’s x,y,z, for respective quantum numbers corresponding
to motions in the x,y,z directions. The allowed energies are:"
"
The number of of conduction electrons per unit volume which is the number of filled states
per unit volume is the same as the number of electrons per unit volume. The Fermi energy is:"
"
We are interested in Aluminum which is 18.1E28 electrons per cubic meter gives a Fermi
energy in electron-volts of 11.7 eV. Aluminum is part of our equation (the dummy in the Si
differential) that produces Stokes theorem:"
"
Or,…"
E
F
−
h
2
2m
e
(
∂
2
∂x
+
∂
2
∂y
+
∂
2
∂z
)
ψ (x, y, z) = E ψ (x, y, z)
ψ (x, y, z) =
2
L
x
sin
n
x
π x
L
x
2
L
y
sin
n
y
π x
L
y
2
L
z
sin
n
z
π x
L
z
π
2
h
2
2m L
2
(
n
2
1
+ n
2
2
+ n
2
3
)
n
e
E
F
=
h
2
2m
e
(
3π
2
N
V
)
2
3
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al)
]
dydz =
∫
Ge
Si
d x
of 14 44
"
"
Where and "
Band Theory
A conductor is different than an insulator in how its electrons respond to an electric
field. In a conductor a large amount of electrons respond whereas electrons in
insulators belong to completely filled bands. We are interested in silicon and
germanium as well and we have only considered aluminum a conductor.
Semiconductors are like insulators except in that they have a small energy gap
between the filled band and the next unavailable unfilled band. The bands at T=0 K are
completely filled in both insulators and semiconductors, however at room temperature
the energy gap is small enough that that electrons can escape from the valence band
to the conduction band and they can respond to an applied field. This band gap in
silicon is 1.14 eV and in germanium is 0.67 eV. In GaAs is 1.43 eV, and in GaP is 2.26
eV at 300K."
Thus, we have if we take gallium arsenide band gap in eV to be the differential As-Ga
across Ge in molar mass…"
From"
"
"
And, this is…"
"
"
"
∫
1
0
∫
1
0
Si
B
(As − Ga)dydz =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al)dydz =
2
3
∫
Ge
Si
dz
1/3 ≈ 1 − ϕ
2/3 ≈ ϕ
∫
1
0
∫
1
0
Si
B
(As − Ga)dydz =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Si
B
(Ga As)dydz =
∫
1
0
∫
1
0
1.14eV
B
(1.43eV )dydz
1
3
∫
Ge
Si
dz =
1
3
(0.67 − 1.14)
(−0.1567eV )B = 1.6302eV
2
B = (10.4eV )
of 15 44
The Boltzmann constant is "
This times room temp in kelvin 300K yields 0.02585eV is the kinetic energy of a particle
at room temperature to get an idea of the order of energies we are discussing. The
Fermi energy for Au=5.53 eV and for Ag=5.49 eV. If we divide the Fermi energy of
aluminum by that of gold or silver we see they are about half that of aluminum. The
interesting thing here is that B=10.40333/2 is 5.20eV is close to
where 8.298eV is the first ionization energy of boron and "
B/2=5.20eV is approximately the activation energy of boron which is 4.7eV the energy
required for a reaction. If we take 2/3 of 8.298eV we have 5.532eV. This averaged with
4.7 eV is 5.116eV is almost exactly our B/2=5.20eV. It makes sense we divide by two
because this is the average between boron band gap and ionization energy which is
1.5eV plus 8.298eV equals 5.6eV is close to 5.20eV. In this sense we see AI as a
mathematical construct in terms of molar mass is in correlation with the quantum
mechanical model of these elements in terms of Fermi energies, band gap energies
and dissociation energies, as well as ionization energies and activation energies. While
the amount of doping agents used to dope semiconductors is extraordinarily small, the
effect is enormous, increasing the conductivity significantly. We have said biological life
describes AI elements in the three by three AI matrix that is conveniently pulled out of
the periodic table, and that this has a dynamic function utilizing boron and aluminum.
But just what are the connections of these mathematical constructs to biological life
elements. We address that here,…"
Silicon and Carbon
We guess that artificial intelligence (AI) has the golden ratio, or its conjugate in its means
geometric, harmonic, and arithmetic by molar mass by taking these means between doping
agents phosphorus (P) and boron (B) divided by semiconductor material silicon (Si) :
Which can be written
k
B
= 8.617E − 5eV ⋅ K
−1
ϕ(8.298eV ) = 5.128eV
PB
Si
=
(30.97)(10.81)
28.09
= 0.65
2 PB
P + B
1
Si
=
2(30.97)(10.81)
30.97 + 10.81
1
28.09
= 0.57
0.65 + 0.57
2
= 0.61 ≈ ϕ
of 16 44
We see that the biological elements, H, N, C, O compared to the AI elements P, B, Si is the
golden ratio conjugate (phi) as well:
So we can now establish the connection between artificial intelligence and biological life:
Which can be written:
Where HNCO is isocyanic acid, the most basic organic compound. We write in the arithmetic
mean:
Which is nice because we can write in the second first generation semiconductor as well
(germanium) and the doping agents gallium (Ga) and arsenic (As):
Where
Where ZnSe is zinc selenide, an intrinsic semiconductor used in AI, meaning it doesn’t require
doping agents. We now have:
PB(P + B) + 2PB
2(P + B)Si
≈ ϕ
C + N + O + H
P + B + Si
≈ ϕ
(P + B + Si )
PB(P + B) + 2PB
2(P + B)Si
≈ (C + N + O + H )
PB
[
P
Si
+
B
Si
+ 1
]
+
2 PB
P + B
[
P
Si
+
B
Si
+ 1
]
≈ 2HCNO
[
PB +
2 PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
≈ 3HNCO
[
PB +
2 PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
≈ HNCO
[
Ga
Ge
+
As
Ge
+ 1
]
Zn
Se
≈
[
P
Si
+
B
Si
+ 1
]
[
Ga
Ge
+
As
Ge
+ 1
]
PB
(
Zn
Se
)
+
2 PB
P + B
(
Zn
Se
)
+
P + B
2
(
Zn
Se
)
≈ HNCO
of 17 44
Germanium And Carbon
We could begin with semiconductor germanium (Ge) and doping agents gallium (Ga) and
Phosphorus (P) and we get a similar equation:
,
In grams per mole. Then we compare these molar masses to the molar masses of the
semiconductor material Ge:
Then, take the arithmetic mean between these:
We then notice this is about the golden ratio conjugate, , which is the inverse of the golden
ratio, . . Thus, we have
1.
2.
This is considering the elements of artificial intelligence (AI) Ga, P, Ge, Si. Since we want to find
the connection of artificial intelligence to biological life, we compare these to the biological
elements most abundant by mass carbon (C), hydrogen (H), nitrogen (N), oxygen (O),
phosphorus (P), sulfur (S). We write these CHNOPS (C+H+N+O+P+S) and find:
A similar thing can be done with germanium, Ge, and gallium, Ga, and arsenic, As, this time
using CHNOPS the most abundant biological elements by mass:
2GaP
Ga + P
= 42.866
GaP = 46.46749
2GaP
Ga + P
1
Ge
=
42.866
72.61
= 0.59
GaP
1
Ge
=
46.46749
72.61
= 0.64
0.59 + 0.64
2
= 0.615
ϕ
Φ
ϕ ≈
1
Φ
GaP(Ga + P) + 2GaP
2(Ga + P)Ge
≈ ϕ
GaP(Ga + P) + 2GaP
2(Ga + P)Si
≈ Φ
CHNOP S
Ga + As + Ge
≈
1
2
[
Ga As +
2Ga As
Ga + As
+
Ga + As
2
][
Ga
Ge
+
As
Ge
+ 1
]
≈ CHNOPS
[
Ga
Si
+
As
Si
+ 1
]
of 18 44
We can also make a construct for silicon doped with gallium and phosphorus:
And for germanium doped with gallium and phosphorus:
Here is a table of the AI biological equations…
Ga As
(
O
S
)
+
2Ga As
Ga + As
(
O
S
)
+
Ga + As
2
(
O
S
)
≈ CHNOPS
O
S
≈
[
Ga
Ge
+
As
Ge
+ 1
]
[
Ga
Si
+
As
Si
+ 1
]
Ga As(G a + As) + 2G a As
2(Ga + As)Ge
≈ 1
C + H + N + O + P + S
Ga + As + Ge
≈
1
2
(C + N + O + H ) ≈
2(Ga + P)Si
GaP(Ga + P) + 2GaP
(P + B + Si )
HNCO ≈
2(Ga + P)Si
(Ga + P)
[
GaP +
2GaP
Ga + P
]
(P + B + Si )
HNCO ≈
2(P + B + Si )Si
GaP +
2GaP
Ga + P
GaP(Ga + P) + 2GaP
2(Ga + P)Ge
≈ ϕ
[
GaP +
2GaP
Ga + P
+
Ga + P
2
][
P
Ge
+
B
Ge
+
Si
Ge
]
≈ HNCO
[
Ga
Ge
+
As
Ge
+ 1
]
GaP
(
B
S
)
+
2GaP
Ga + P
(
B
S
)
+
Ga + P
2
(
B
S
)
≈ HNCO
of 19 44
The Fundamental AI Bioequations
[
PB +
2 PB
P + B
+
P + B
2
][
P
Si
+
B
Si
+ 1
]
≈ HNCO
[
Ga
Ge
+
As
Ge
+ 1
]
[
Ga As +
2Ga As
Ga + As
+
Ga + As
2
][
Ga
Ge
+
As
Ge
+ 1
]
≈ CHNOPS
[
Ga
Si
+
As
Si
+ 1
]
[
GaP +
2GaP
Ga + P
+
Ga + P
2
][
P
Ge
+
B
Ge
+
Si
Ge
]
≈ HNCO
[
Ga
Ge
+
As
Ge
+ 1
]
HNCO ≈
2(P + B + Si )Si
GaP +
2GaP
Ga + P
PB(P + B) + 2PB
2(P + B)Si
≈ ϕ
Ga As(G a + As) + 2G a As
2(Ga + As)Ge
≈ 1
GaP(Ga + P) + 2GaP
2(Ga + P)Ge
≈ ϕ
GaP(Ga + P) + 2GaP
2(Ga + P)Si
≈ Φ
C + N + O + H
P + B + Si
≈ ϕ
C + H + N + O + P + S
Ga + As + Ge
≈
1
2
Zn
Se
≈
[
P
Si
+
B
Si
+ 1
]
[
Ga
Ge
+
As
Ge
+ 1
]
O
S
≈
[
Ga
Ge
+
As
Ge
+ 1
]
[
Ga
Si
+
As
Si
+ 1
]
of 20 44
Using The Fundamental Equations
Now that we have outlined the fundamental AI Bioequations, let us put them to use. We
consider:
Making the approximations: ,
, we obtain:
Which can further be written by saying :
Which is interesting because the Si times itself is then equal to something times itself in that Ge
and Si are both semiconducting materials, but Ge is larger than Si, however this is compensated
for by reducing it by a factor of the golden ratio conjugate, phi. The equation is however only
79% accurate because there has been a lot of drift due to so many approximations. However if
we reduce phi by a factor of itself and write:
It is then 99% accurate:
If we do the same with the other and write:
(P + B + Si)
PB(P + B) + 2PB
2(P + B)Si
≈ (C + N + O + H )
HNCO ≈
2(P + B + Si)Si
GaP +
2GaP
Ga + P
GaP ≈ ϕGe
2GaP
Ga + P
≈ ϕG e
PB ≈ ϕSi
2Si
2
Ge
= ϕSi +
2P B
P + B
2P B
P + B
≈ ϕSi
Si
2
= ϕGeSi
Si
2
= ϕ
2
GeSi
28.09 = (72.64)(28.09)(0.381924) = 27.9g/m ol
27.916
28.09
= 0.99
2Si
2
Ge
= ϕ
2
Si +
2P B
P + B
of 21 44
We have:
Which is better but still only 81% accurate. However if we write it:
Then it is 95.87% accurate. But we see in the first approximation that . That is we
have boron, the element that is out of place in the AI periodic table resulting in the dynamics of
our equations. So, we can write…
This gives…
Which is 26.836 which is close to aluminum (Al=26.98) which is the dummy representative for
boron in our equations. We have incredibly:
With an accuracy of nearly 100%. This becomes…
While phosphorus, boron, silicon, and germanium and gallium and arsenic are the primary AI
elements, gold (Au), Silver (Ag) and copper (Cu), are the fundamental AI elements in that they
conductive, ductile, and malleable. Incredibly, the number 3 in the above equation is the ratio of
gold to copper in molar mass, so we have…
21.72 = (0.381924)(28.09) + 16.026 = 10.72 + 16.026 = 26.75
2Si
2
Ge
= ϕ
3
Si +
2P B
P + B
phi
2
Si ≈ B
2Si
2
Ge
= B +
2P B
P + B
10.81 + 16.02 = B +
2P B
P + B
Al = B +
2P B
P + B
Al = B
3P + B
P + B
Al = B
Au
Cu
P + B
P + B
Au
Cu
=
196.97
63.55
= 3.099 ≈ 3
of 22 44
Conclusion
Since we have"
"
"
"
And "
"
"
We have"
"
"
We also have that"
"
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al)
]
dydz =
∫
Ge
Si
d x
∫
1
0
∫
1
0
Si
B
(As − Ga)dydz =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al)dydz =
2
3
∫
Ge
Si
dz
C + N + O + H
P + B + Si
≈ ϕ
PB(P + B) + 2PB
2(P + B)Si
≈ ϕ
∫
1
0
∫
1
0
Ge
B
(P − Al)dydz =
C + N + O + H
P + B + Si
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al)dydz =
PB(P + B) + 2PB
2(P + B)Si
∫
Ge
Si
dz
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
of 23 44
Which is a form of Maxwell’s equation:"
"
Where"
"
Making"
"
Become"
"
Because"
"
Where Ge/Si are not equal to mu times epsilon not, but play the role of that product.
The golden ratio conjugate plays the role of c the speed of light. This gives the
halfwave"
"
And we have said"
"
This is interesting because the Fermi energy for gold (Au) is 5.53eV and for silver (Ag) is
5.49 eV. This is on the order of our B/2=5.20eV. Gold and silver are the best electric
conductors and as such are used in AI prolifically. Copper is a very good conductor as
∇ ×
B = μ
J + μϵ
0
∂
E
∂t
Ge
Si
= μϵ
0
(
∇
2
−
1
c
2
∂
2
∂t
)
E = 0
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
Si = 0
c =
1
ϵ
0
μ
≈ ϕ
y(x) = e
B
Al
x
+
Ag
Cu
Al = B
Au
Cu
P + B
P + B
of 24 44
well but is used the most because it is cheap and abundant. I believe this relevant in
light of "
"
"
The Fermi energy of Aluminum (Al) is 11.7eV that of copper (Cu) is 7.00 eV. The fermi
energies of Au and Ag added together are that of Al."
Au+Ag=Al=5.49+5.53=11.02eV~11.7eV"
Semiconductor Devices
Coulomb’s Law:"
"
Which states the force exerted between two charges q1 and q2 are directly
proportional to the product of their charges, and inversely proportional to the square of
the distance between them. "
The charge of a proton is the same as the charge of an electron, except for the proton
the charge is positive and for the electron it is negative. Like charges repel and
opposite charges attract. The charge of a proton and of an electron is .
The constant of proportionality in Coulomb’s Law is:"
"
The equation of force of a charge moving in an electric field and magnetic field is
given by the Lorenz Force:"
"
The Hall effect is the production of a potential due to the motion of a conductor in a
magnetic field. The Hall voltage ( ) created by the drift velocity v of an electron in a
magnetic field B through a conductor of width w is#
#
"
y(x) = e
B
Al
x
+
Ag
Cu
Al = B
Au
Cu
P + B
P + B
F = k
e
q
1
q
2
r
2
1.6 × 10
−19
C
k
e
=
1
4πϵ
0
= 9 × 10
9
E
B
F = q(
E +
v ×
B )
V
H
V
H
= vBw
of 25 44
An n-type semiconductor is created by doping a semiconductor material such as
silicon valence 4- with a doping agent or, donor, of an extra electron such as arsenic
valence 5-. Such a doped substance has its electrons easily excited from the valence
band to the conduction band by a small amount of heat. Easily because the extra
electron has no positive charges to keep it in. As these electrons move into the
conduction band, they leave behind holes which can be thought of as a current as well
because they have electrons falling into them. This current is said to be carried by
minority carriers, and the current due to electrons in the conduction band is said to be
that of majority carriers. In a p-type semiconductor the majority carries are said to be
the free holes because in this case the material is doped with something like gallium
valence -3 which one electron less than the silicon valence 4-. Minority carriers are
then holes left behind due to thermal excitation of electrons in the valence band across
the gap to the conduction band."
If we put a semiconductor strip in a magnetic field (field lines at right angles to its
surface) it drives positive holes to its upper end which gather there until the electric
field acting along the surface (right angles to the magnetic field) balances with the force
due to the magnetic field:"
"
"
And we have "
The drift velocity v, of the majority carriers can thus be found and is:"
"
Or we can get it from the Hall voltage. From this we can determine the current density j
as well:"
"
Where is the number of charge carriers per unit volume, q is the charge of an electron,
and v is the drift velocity. The flow of electrons into the conduction band equals the
holes left behind, or n=p"
Semiconductor devices like diodes and transistors are created when we connect n-
type semiconductors with p-type semiconductors. Thousands, even millions of these
device can be formed on a small piece of silicon with connections between the
conducting paths."
F
B
= qvB
F
E
= qE
E = vB
v =
E
B
j = nq v
of 26 44
For diodes n-type and p-type semiconductors are joined with negative donors on the p
side and positive acceptors on the negative side at the p-n junction. In the reverse bias
configuration the + side of a battery is applied to the n-type side of the diode which
widens the depletion layer increasing the potential energy across the p-n junction. As
such few electrons get through to establish a current. However if the + side of the
battery is connected to the p-type material the depletion layer is narrowed, the
potential across the junction decreases and electrons flow from one side to the other.
This makes the diode a one-way valve."
The fraction of electrons that have enough energy to diffuse across the potential barrier
is given by the Maxwell-Boltzmann distribution. In the reverse bias hookup it is:"
"
And much better in the forward bias connection with voltage hooked up in that bias:"
"
is the current with no voltage applied across the diode, just the minority carriers,
which is independent of the bias voltage. Just the flow of electrons from the valence
band to the conduction band due to heat. We have the net current is:"
"
To make a transistor we join three semiconductors so we can have npn or pnp
transistors. These are three lead diodes. I if we apply a voltage to the the middle p in
the npn transistor, to the base, we have a junction transistor which is not just a one
way valve like the diode, but that can be turned on or off to control the current flow.
Hooking up a positive terminal to the base causes a small current to flow which in turn
causes a large current to flow at the collector (n) from the emitter (the other n). By
controlling the base current we can control the collector current. The transistor is like a
valve and can be used to either amplify currents with a continuous analog signal or
switch currents either on or off so we can make switches that encode in binary making
computers possible, on or off, 1 or zero."
The Mathematical Connection
We can greatly simply the problem if we want to search for an explanation for AI as a
mathematical construct by looking at diodes instead of transistors. While logic gates
are built so we can have AI by forming millions of interconnected transistors on silicon
crystals and interconnecting them, we can build all of the logic gates with diodes; the
AND gate, OR gate, NOT gate…"
Ne
eV/k
B
T
V
B
I = I
0
e
eV
B
/k
B
T
I
0
I
net
= I
0
(
e
eV
B
/k
B
T
− 1
)
of 27 44
Indeed early on, in first developing AI we made the equivalent of diodes with vacuum
tubes. They had the same elements and principles behind them. In this case they were
tungsten filaments in a glass (SiO2) tube. Electrons emitted by thermionic emission of
heating the filament is like the emitter in a np diode, is the cathode. Electrons are
attracted by a metal plate, the anode like the collector in the np diode. In a triode we
have a metal screen or grid placed between the cathode and anode which can control
the current by voltage applied to it like the base in a npn transistor."
We are looking for some characteristics that characterize germanium and silicon
diodes, if we want explain our equations that characterize them as mathematical
constructs. We begin like this…"
In the reverse bias configuration for extreme values of reverse bias the material ionizes
causing an avalanche of current which occurs at breakdown voltage. The current is
this the reverse saturate current."
The Hall effect is best described as the electric field E that arises if a metal or
semiconductor carrying a current I is placed in a magnetic field of flux density B which
is developed along a direction perpendicular to B and I."
The values we are working with are:"
=diode current"
=diode reverse saturation current"
V=External voltage applied to diode"
=a constant 1 for germanium 2 for silicon"
is the Thermal Voltage"
=Boltzmann Constant is 1.380649E-23 J/K"
charge of electron (1.60217662E-19C)"
T=temperature of diode junction in degrees kelvin"
Room Temperature 300K is "
Our equation is:"
"
I
0
−I
sat
I
I
0
ξ
V
T
= K T/q = T/11600
k
B
q =
V
T
= 26mv
I = I
0
(
e
(V/ξV
T
)
− 1
)
of 28 44
…"
The one value that is constant for both Si and Ge diodes is the their forward biases, the
voltage that turns them on. For Si this is 0.7V for Ge this is 0.3V. We say"
"
"
T=6960 degrees Kelvin (12068.33 degrees F)"
T=3480 degrees Kelvin (5804.33 degrees F)"
At these temperatures the currents generated in Si and Ge diodes without being
connected to a voltage could turn on another set of Si and Ge diodes."
We can also write "
So we have"
"
"
Silicon: "
Germanium: "
Thus if V over V thermal is in the ratio of Silicon forward bias to germanium forward
bias we have factors of 1.718 and 6.389 for constants of Si=2 and Ge=1,
respectively. We see that gold, the heaviest fine conductor divided by germanium the
heaviest fine semiconductor is"
(196.97g/mol)/(72.64g/mol)=2.71~e=2.718"
And, the second heaviest fine conductor, silver, divided by silicon the second heaviest
fine semiconductor is"
(107.87g/mol)(28.09g/mol)=3.840~e+1=3.718"
0.6V = K T/q = T/11600
0.3V = K T/q = T/11600
V/V
T
= 0.6V/0.3V
I
I
0
= e
2/2
− 1
I
I
0
= e
2/1
− 1
I
I
0
= e − 1 = 1.718
I
I
0
= e
2
− 1 = 6.389
ξ
of 29 44
"
"
The lightest fine conductor is copper. As we progress from germanium, to silicon, the
next, and lighter element in that group is carbon, which though is not a fine
semiconductor, it is the core element to biological life. We have:"
(63.55g/mol)/(12.01g/mol)=5.2914"
"
It is at this point that we remember"
"
We bring this up because Boron out of place in the AI periodic table being with the
biological elements, aluminum in its place use as the dummy differential across Si,
could be substituted with gallium (Ga) even though it belongs to the differential across
Ge because it is symmetrical placed with boron (B) around aluminum (Al). We have"
"
This is in our halfwave"
"
That comes from"
"
Where"
I
I
0
=
Au
Ge
− 1
I
I
0
=
Ag
Si
− 2
I
I
0
=
Cu
C
− e = 2.57 ≈
Ga
Al
=
69.72
26.98
= 2.584
Si
B
(As − Ga) +
Ge
B
(P − Al) =
2SiGe
Si + Ge
B
Al
=
10.81
26.98
= 0.40
y(x) = e
B
Al
x
+
Ag
Cu
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
Si = 0
of 30 44
"
The Broad Picture
We see as we progress from the parallel placements (Au/Ge) through the table to (Ag/
Si) and finally to (C/Cu) where we have run out of slots, we progress from AI to AI and
finally Life (C/Cu). But what does the fine electric conductor of AI, which is Cu, have to
do with carbon and biological life?"
The biochemistry of all terrestrial life seems to be similar throughout the spectrum of
species, and so much so that it would seem we all descended from a common
ancestor. Of course at the most fundamental level life as we know it is based on
carbon. This is because it has 4 valence electrons allowing it to combine in long chains
with hydrogen called hydrocarbons, and combine readily with oxygen 2-, and nitrogen
3-. Biologist have hazarded to look at silicon as a candidate since it has a valence of 4
as well, and came to the conclusion that life did not form like this because in the
presence of oxygen it readily combines with it making SiO2 silicon dioxide the basic
ingredient of sand. Perhaps many different organisms originated differently and
independently but at some point we may have a common ancestor. Her name is LUCA
(Last Universal Common Ancestor). Using a comparative approach biochemists have
traced our ancestry back to LUCA who shared the traits we have today. Some of these
traits are she stored her genetic information in DNA, and made use of the same twenty
amino acids we use to make our proteins with the same RNA machinery and genetic
code we use."
However, in dating LUCA we looked at the metabolic pathways common to all of life
and found her metabolism was based on iron (Fe). In today’s oxygen rich environment
iron quickly oxidizes to its ferric state and is highly insoluble. It is believed then that
LUCA lived before the earth was rich in oxygen. Today we use in our metabolic
pathways copper (Cu) because its oxidized state is more soluble than its reduced
forms. Marine organisms use iron and it is the limiting nutrient in the ocean and the
marine organisms have developed mechanisms with which to to extract iron form
bacteria. But the important point here is that the earth became oxygen rich with the
arrival of photosynthesizers, plants that could use energy from the sun to make
electrochemical energy (sugars) from carbon dioxide and in the process make oxygen.
The oxygen rich atmosphere seems to have happened about 2 billion years ago but the
evidence for life on earth goes back as far as 3.5 billion years, putting LUCA’s age in a
wide gap of 2 billion to 3.5 billion years old.$
ϕ = 0.618 =
1
Ge /Si
=
1
2.58597
= 0.62185
of 31 44
Divergence
We did a stokes theorem formulation of the AI elements. We want to do a divergence theorem
formulation. Our stokes theorem formulation was"
"
We want to consider"
"
We have to find the F-vector. It is from"
"
, , "
"
"
"
We have:"
"
We have said the speed of light in electrodynamics is given by"
"
∫
1
0
∫
1
0
Si
B
(As − Ga)dydz =
1
3
∫
Ge
Si
dz
∫∫∫
V
∇ ⋅
F dV =
∫∫
S
(
F .
n)d
S
i
j
k
∂
∂x
∂
∂y
∂
∂z
0
Si
B
Ga ⋅ z
Si
B
As ⋅ y
=
Si
B
(As − Ga)
i
F
x
= 0
F
y
=
Si
B
Ga ⋅ z
F
z
=
Si
B
As ⋅ y
∇ ⋅
F = 0
∫
V
0dV = constant
∫∫
S
(
F
x
, F
y
, F
z
)
⋅
(
dydz, d xdz, d xdy
)
=
Si
2B
(
Ga ⋅ z
2
+ As ⋅ y
2
)
Si
2B
(
Ga ⋅ z
2
+ As ⋅ y
2
)
=
Q
ϵ
0
c =
1
ϵ
0
μ
of 32 44
And in our formulation of the AI elements plays the role of the speed of light, is the
golden ratio conjugate is:"
"
Thus"
"
"
Thus we see"
"
"
Thus in our formulation the golden ratio conjugate plays the role of the speed of light in
electrodynamics. This makes sense because in relativity theory velocities are
considered a fraction of the speed of light because the speed of light is the fastest
speed attainable. The speed v squared over the speed of light squared is subtracted
from 1. That is the Lorentz contraction is:"
"
See the following storyboard pages to the computations…"
ϕ
ϕ =
1
Ge /Si
Ge /Si = μϵ
0
ΔS
Si
= μ =
Ge − Si
Si
= Φ
ϵ
0
=
Ge
Ge − Si
= Φ
1
μϵ
0
=
1
Φ
2
= ϕ
L = L
0
1 −
v
2
c
2
of 33 44
$
of 34 44
$
of 35 44
$
of 36 44
$
of 37 44
$
of 38 44
We see the eccentricity of the ellipse "
"
Plays the role of the beta factor in"
"
In the equation"
"
We see the eccentricity is"
"
Which is 9 figures eight of which are the first eight consecutive integers 1,2,3,4,5,6,7,8.
The only number that occurs twice is 2. Kind of interesting."
1 −
D
C
L = L
0
1 −
v
2
c
2
Si
2B
(
Ga ⋅ z
2
+ As ⋅ y
2
)
=
Q
ϵ
0
1 −
Ga
As
= 1 −
69.72
74.92
= 0.263452781
of 39 44
The Heart of the Matter
If we take our halfwave:"
"
And apply the schrodinger equation:"
"
Then,…"
"
"
We know that"
"
Or,…"
"
We notice"
"
And"
"
This says"
"
The molar masses of Ge and Si are such that the geometric mean between Ge and Si is the
difference between Ge and Si. This has solutions:"
$
y(x) = e
B
Al
+
Ag
Cu
(
∂
2
∂x
−
1
ϕ
2
∂
2
∂x
)
(
e
B
Al
x
)
= 0
(
B
Al
)
2
(
1 −
1
ϕ
2
)
= 0
ϕ = 1
ϕ =
1
Φ
ϕΦ = 1
Ge
Ge − Si
=
72.64
72.64 − 28.09
= 1.63 ≈ Φ
Si
Ge − Si
=
28.09
72.64 − 28.09
= 0.63 ≈ ϕ
GeSi = Ge − Si
y =
1
2
(3x − 5x)
y =
1
2
(3x + 5x)
of 40 44
If x=1 we have:"
0.381, 2.618"
If x=2 we have"
0.76, 5.236"
If x=3 we have"
1.145898, 7.854"
If x=4 we have"
1.52786, 10.47"
If x=5 we have"
1.9098, 13.090"
This is a pair of intersecting lines:"
"
"
With slopes"
"
"
The plots are…$
y = (1 − ϕ)x = 0.38x
y = (1 + Φ)x = 2.62x
m
1
= (1 − ϕ)
m
2
= (1 + Φ)
of 41 44
$
of 42 44
Since"
"
"
But we have said:"
"
But we also have"
"
This says"
"
"
Or…"
0=1"
Since we have said"
"
Is"
"
Then we might guess"
"
Is 1 is E and is c and is m in"
1 − ϕ =
Si
Ge
=
28.09
72.64
= 0.3867
1 + Φ =
Ge
Si
=
72.61
28.09
= 2.58597
Ge
Si
=
1
ϕ
1 − ϕ =
Si
Ge
Φϕ
2
= 1
1
ϕ
(ϕ − 1) = 1
1 −
Ga
As
1 −
v
2
c
2
Φϕ
2
= 1
ϕ
Φ
of 43 44
"
But that 0=1 may be that the quantum states 0 and quantum states 1 are
interchangeable in Hund’s rule and allow for more than 2 electrons in a bond. And we
see Alexander got to the heart of the matter when he wrote concerning my earlier work
in this project:"
Александр Сергеевич (Alexander) of Institute of Physical Chemistry (Moscow) wrote "
Your approach is very relevant and touches on one of the main provisions of quantum
chemistry-namely, the well-known Hund’s rule that sets the number of electrons
involved in the formation of a bond (usually n=2). However, it is still not clear why only
two electrons must necessarily participate in the formation of a chemical bond - there
are no prohibitions in quantum mechanics in this case! Perhaps your proposed
classification confirms the possibility of participation in the formation of a chemical
bond of several electrons, more than two in number! Then it is possible to substantiate
the presence of some criterion that makes it possible to distinguish between
biologically active and passive elements. And this is extremely important for
understanding how life arose on Earth!!
E = mc
2
of 44 44
The Author